11.1 What is curvature?
11.2 Tidal forces
11.3 Riemann curvature 121
11.4 Symmetries of the Riemann tensor of Rieman
11.5 The Ricci tensor and Ricci scalar
11.6 Example computations 128 11.7 Geodesic deviation revisited
Chapter summary 130
Exercises ^(1){ }^{1} Our problem in finding the curvature of spacetime is more acute, of course, since we can't use any method analogous to noting the change in elevation of distant stars to work out the nature of the curvature. ^(2){ }^{2} G. F. Bernhard Riemann (1826-1866) ^(3){ }^{3} The motion of particles in curved spacetime is often described in terms of a rubber sheet model. This involves picturing space (in two dimensions) as being akin to an elastic sheet. Massive objects make indentations in the sheet and so all objects have their motion affected by having to negotiate these indentations as they move around. Although useful as a picture, it should also be kept in mind that this model describes the curvature of space and not spacetime. In some cases (such as the precession of the orbit of a planet around a star), the metric component describing the timelike variable has a larger effect on the corrections to Newtonian motion than those relating to the spatial coordinates. The combined effect is therefore certainly a spacetim one. This is another manifestation one. This is another manifestation of the feature that space and time are inextricably linked in relativity.
From the point of view of general relativity, humans are condemned to be trapped in spacetime. We cannot, for example, lift ourselves out of spacetime and look at its structure from the outside, by embedding it in some higher dimensional space. If gravitation is manifested in terms of the curvature of spacetime then how are we to measure, understand and explain this curvature? Of course humans have experienced a similar problem before: the surface of the Earth is a two-dimensional space and there are several means of working out that the Earth is, to a good approximation, a spherical ball. ^(1){ }^{1}
Carl Friedrich Gauss made some of the most significant progress in finding that there was a way to convert measurements of distances, made by an observer trapped on a two-dimensional surface, into an objective description of the curvature of that space. However, Bernhard Riemann ^(2){ }^{2} was engaged in an even more ambitious programme to evaluate the curvature of higher dimensional spaces. He solved the problem in 1854 and introduced a mathematical description that would be built up by Christoffel, Ricci, Levi-Civita and others. It is this description that was incorporated into physics by Einstein. Riemann's description of curvature is encoded into a tensor R\boldsymbol{R}. We shall see that a non-zero R\boldsymbol{R} is the sure-fire way of telling mathematically that a spacetime is truly curved. ^(3){ }^{3}
11.1 What is curvature?
Euclid's fifth postulate, known as the parallel postulate, may be paraphrased as saying that parallel lines always remain parallel. Euclid did not prove this, which is why it remained a postulate. It turns out only to be the case in flat space.
The surface of a cylinder appears curved, but this curvature is extrinsic: we can recreate (or wrap) the curved surface of the cylinder using a flat piece of paper with two of its opposite edges identified. We can therefore unravel the cylinder, turning it into a flat plane. Parallel lines on a cylindrical surface never meet, therefore, as they are equivalent to parallel lines on the flat plane. The sphere is different, its curvature is intrinsic: it is impossible to wrap the sphere in a flat piece of paper in the way that we did the cylinder. Parallel lines on the surface of a sphere don't remain parallel; they eventually meet.
Curved surfaces can be classified in terms of the paths of such parallel lines. On a flat surface parallel lines remain parallel. On a surface with positive curvature initially parallel lines converge; on a surface with negative curvature they diverge (Fig. 11.1). An example of a surface with positive curvature is the surface of a sphere; surfaces with negative curvature can be thought of as being saddle-like.
There are two simple ways of telling whether curvature is intrinsic or extrinsic. The first, motivated by the discussion so far, is to set two free, spatially separated particles moving parallel. If the space is intrinsically curved, the particles' paths will deviate from parallelism. This effect, shown in Fig. 11.1 is known as geodesic deviation.
The second method is to parallel transport a vector in a closed loop. If the space has intrinsic curvature, then the vector will rotate. If the space has extrinsic curvature, there will be no change in direction. The rotation is most easily seen by examining parallel transport on the surface of a sphere as shown in Fig. 11.2.
In summary, to measure the amount of curvature present we choose one of the two methods: (i) identify a fiducial geodesic and compare it to another geodesic whose tangent vector was originally parallel. The extent of the deviation tells us the amount by which the space is curved. (ii) Take a vector and parallel transport it around a loop in spacetime. The amount by which the vector changes its direction provides a measure of curvature. As we shall show, these two methods both provide the same measure of curvature: the components of the Riemann curvature tensor R\boldsymbol{R}.
The ultimate source of curvature is the metric field. We saw in the previous chapters how the connection coefficients were derived from the first derivatives of the components of the metric (with the rule of thumb expression that del g)rarr( bar(Gamma))\partial g) \rightarrow(\bar{\Gamma}) ). The Riemann curvature tensor is built from the first and second derivatives of the metric and so we have the rule of thumb
This leads us to expect that the components of this tensor can also be written in terms of the connection coefficients and their derivatives (i.e. ( bar(Gamma))+(del Gamma_)rarr(R_)(\bar{\Gamma})+(\underline{\partial \Gamma}) \rightarrow(\underline{R}), which, as we shall see, turns out to be the case.
11.2 Tidal forces
The simplest gravitating system might be expected to have a static metric, completely independent of time. This might represent the geometry around a star. Such a geometry would allow us to identify a static frame of reference (i.e. the rest frame of the star). We could then perform experiments in which we placed a single particle at rest in this frame and watched to see if it started accelerating. This would allow us to identify a gravitational field. However, this is not a good description of our Universe, where the principles of relativity teach us that it is not generally possible to identify a static frame against which to identify accelera-
Euclid's postulates are:
I: A straight line segment can be drawn joining any two points.
II: Any straight line segment can be extended indefinitely in a straight line. III: Given any straight line segment, a circle can be drawn having the segment as radius and one endpoint as centre. IV: All right angles are congruent. V: If two lines are drawn which intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect the two lines inevitably must intersect
each other on that side if extended far each other on that side if extended far
enough. This postulate is equivalent to the parallel postulate.
Fig. 11.1 Geodesic deviation. In a curved spacetime, particles fall along geodesics that are parallel at some point in spacetime. Extending the point in spacetime. Extending the paths into the future and past, the
world lines of the particles meet if the space is positively curved.
Fig. 11.2 Transporting a vector around a closed loop on the surface of a around a closed loop on the surface of a
sphere. The vector rotates as a result of the intrinsic curvature of the spherical surface. ^(4){ }^{4} We prove this statement about the volume staying constant at the end of this section.
Fig. 11.3 Tidal forces. (a) The forces on a cloud of particles surrounding an observer that are falling towards the Earth, shown in the Earth's rest frame. (b) The same forces in the rest frame of the observer.
Fig. 11.4 Particles falling in a Newtonian potential, with separation vector vec(s)\vec{s}.
tion. We saw that taking relativity into account leads to the equivalence principle, telling us that from the motion of a single particle we cannot tell the difference between acceleration due to a choice of coordinates and acceleration due to spacetime curvature. The best we can do, is to examine the relative motion of two particles.
Let's briefly return to the world of Newtonian gravity and discuss geodesic deviation in this limit. Consider a spherical cluster of test particles falling towards the earth. Each particle moves on a straight line through the Earth's centre, but those that are closer fall faster than those further away, owing to the greater force on the closer particles. As a result of the motion, the sphere won't remain spherical, but will be distorted into an ellipse of the same volume. ^(4){ }^{4} In the rest frame of an observer falling with the particles, the forces that cause the change to the shape of bodies in a gravitational field are known as tidal forces. They are illustrated in Fig. 11.3.
Example 11.1
Consider a cloud of dust particles that are initially spherical, that are released in the vicinity of a gravitating mass. Newton's second law gives us the equation of motion for each particle in a gravitational potential Phi( vec(x))=-GM//| vec(x)|\Phi(\vec{x})=-G M /|\vec{x}| (temporarily relaxing our rules on balanced up and down tensor components)
We imagine the trajectories of two particles in the cloud, shown in Fig. 11.4, with a vector vec(s)(t)= vec(y)(t)- vec(x)(t)\vec{s}(t)=\vec{y}(t)-\vec{x}(t) stretching between them. (The connecting vector is assumed small compared to the distance from the gravitating mass in what follows.) We can take the difference in their equations of motion
Define a tensor with components R^(i)_(j)=R_(ij)=(del^(2)Phi(( vec(x))))/(delx^(i)delx^(j))R^{i}{ }_{j}=R_{i j}=\frac{\partial^{2} \Phi(\vec{x})}{\partial x^{i} \partial x^{j}}, leading to the equation of motion of the components of the deviation vector
Let's examine the case where the particle starts with vec(x)=(0,0,z)\vec{x}=(0,0, z), and falls towards a planet, whose centre is the origin. The tensor R^(i)_(j)R^{i}{ }_{j} may be evaluated, giving
The cloud becomes elongated along zz, because (d^(2)s^(z))/((d)t^(2))=2(GM)/(r^(3))z\frac{\mathrm{d}^{2} s^{z}}{\mathrm{~d} t^{2}}=2 \frac{G M}{r^{3}} z. The cloud is compressed along xx and yy, this is because (d^(2)s^(i))/((d)t^(2))=-(GM)/(r^(3))*s^(i)\frac{\mathrm{d}^{2} s^{i}}{\mathrm{~d} t^{2}}=-\frac{G M}{r^{3}} \cdot s^{i}, for i=xi=x and yy.
After this discussion of Newtonian mechanics, we turn to relativistic gravitation. We use the same strategy, but this time using the Riemann tensor R\boldsymbol{R}. The technique is, once again, to assess the geodesic deviation. We identify some freely falling particle on a fiducial geodesic and note its 4 -velocity u\boldsymbol{u}. We compare the geodesic of a different freely falling particle on a different geodesic, whose velocity was originally parallel to the fiducial one. The separation of the two geodesics is xi\boldsymbol{\xi}. If there's curvature present then xi\boldsymbol{\xi} will accelerate. The equation of motion for the separation vector is ^(5){ }^{5}
The (1,3) tensor R(,,\boldsymbol{R}(,,,)isthekeyobjectforassessingcurvature.) is the key object for assessing curvature. Filling the latter three slots with vectors as shown in eqn 11.6 gives rise to a (1,0)(1,0) object (i.e. a vector) that measures the relative acceleration of the geodesics. If R=0\boldsymbol{R}=0 (that is, its components vanish) then space is flat. If the components aren't zero then we are able to say that the space is curved. ^(6){ }^{6}
We shall return to the use of geodesic deviation at the end of the chapter. As a measure of curvature it is conceptually clear, but computations rely on some mathematical tricks we haven't yet encountered. In the next section we make use of the other approach, parallel transport around a loop, in order to obtain an explicit form of the tensor R\boldsymbol{R}.
Example 11.2
Proof that the volume of the cluster of particles is constant as it deforms. Recall that we can write Coulomb's law for the electrostatic force between charges and justify Gauss' law for the divergence of the vec(E)\vec{E}-field from charge density rho_(c)\rho_{\mathrm{c}}, which reads vec(grad)* vec(E)=rho_(c)//epsi_(0)\vec{\nabla} \cdot \vec{E}=\rho_{\mathrm{c}} / \varepsilon_{0}. In the same way, we can write a differential equation relating the gravitational field ^(7) vec(g){ }^{7} \vec{g} to a density of matter rho\rho, which reads vec(grad)* vec(g)=-4pi G rho\vec{\nabla} \cdot \vec{g}=-4 \pi G \rho. In integral form, this is written as
{:(11.7)int vec(g)*d vec(S)=-4pi GM",":}\begin{equation*}
\int \vec{g} \cdot \mathrm{~d} \vec{S}=-4 \pi G M, \tag{11.7}
\end{equation*}
where MM is the total mass in a volume bounded by the closed surface with area vector vec(S)\vec{S}. In words this expression says that the strength of the gravitational field is given by the mass inside the surface vec(S)\vec{S}. If, instead of vec(g)\vec{g} we use the potential Phi\Phi, via vec(g)=- vec(grad)Phi\vec{g}=-\vec{\nabla} \Phi, then we obtain Poisson's equation ^(8){ }^{8} for gravitational fields,
{:(11.8) vec(grad)^(2)Phi=4pi G rho.:}\begin{equation*}
\vec{\nabla}^{2} \Phi=4 \pi G \rho . \tag{11.8}
\end{equation*}
We can use this expression to justify the claim that a sphere of test particles falling towards the earth does not change its volume. Note first that grad^(2)Phi\nabla^{2} \Phi is the sum of eigenvalues of the matrix (del^(2)Phi)/(delx^(i)delx^(3))\frac{\partial^{2} \Phi}{\partial x^{i} \partial x^{3}} that we considered above. We can then use the intervals s^(i)s^{i} from the previous example to build a small sphere of particles. We then have, as a result of our expression ms^(i)=-(del^(2)Phi)/(delx^(i)delx^(j))s^(j)m s^{i}=-\frac{\partial^{2} \Phi}{\partial x^{i} \partial x^{j}} s^{j}, that the second time derivative of the volume of the sphere of particles is determined by the density rho\rho of matter inside the sphere. Therefore, a sphere of free particles falling towards the earth forms an ellipsoid of the same volume as the original sphere, as there is (always) no gravitating mass inside the sphere. (If, on the other hand, we built a large sphere around the earth, we would expect the sphere to reduce its volume as the particles fell towards the earth.) ^(5){ }^{5} This equation is discussed in detail in Chapter 35 where it is justified more thoroughly and compared against the parallel transport method discussed in the chapter. ^(6){ }^{6} The various methods of measuring curvature are discused in detail in curvature are discussed in detail in Chapter 30 . As a rough measure, we
say that the amount of curvature induced by the non-zero Riemann tensor duced by the non-zero Riemann tensor
is, in the orthonormal frame, of order is, in the orth (1//R^( hat(mu))_( hat(nu) hat(alpha) hat(beta))^((1)/(2)))^((1)/(2))\left(1 / R^{\hat{\mu}}{ }_{\hat{\nu} \hat{\alpha} \hat{\beta}}^{\frac{1}{2}}\right)^{\frac{1}{2}}. ^(7){ }^{7} Defined by vec(F)=m vec(g)\vec{F}=m \vec{g}, where vec(F)\vec{F} is the gravitation force on a particle with mass mm.
11.3 Riemann curvature
Although geodesic deviation provides a physical and intuitive means of working out if space is curved, there is a mathematically simpler ^(9){ }^{9} method of determining the form of the Riemann tensor, involving the parallel transport of a vector. Recall that parallel transport is a method to work out how a vector changes due to the change in the underlying coordinate system. This isn't curvature necessarily, we may just be using a hopelessly complicated coordinate system. In order to work out
Fig. 11.5 Transporting a vector around an infinitesimal closed loop. if a spacetime is truly curved, we recall the fate of a vector parallel transported on a spherical surface from Fig. 11.2. We can see from this figure the effect of parallel transportation around a closed loop: the vector rotates. This is a smoking gun, telling us that the surface is truly curved. If it were not, the vector would point in the same direction after parallel transport. In order to assess the curvature of a spacetime we shall follow the same procedure and move a 4 -vector in an infinitesimal loop to see if its direction changes. If it does, we will be able to identify the intrinsic curvature of the spacetime.
So let's be specific and take a vector V\boldsymbol{V} and move it round a closed, infinitesimal parallelogram formed by a vectors delta ae_(1)\delta a \boldsymbol{e}_{1} and delta be_(2)\delta b \boldsymbol{e}_{2}, where delta a\delta a and delta b\delta b are (small) constants (see Fig. 11.5). The idea is that we track the change in a component of V\boldsymbol{V} as follows:
{:[deltaV^(alpha)=((" Change in "V^(alpha)" transported along "delta ae_(1),)/(" then "delta be_(2)," then "-delta ae_(1)," then "-delta be_(2)))],[(11.9)=((" Measure of the curvature ")/(" of spacetime "))_(mu)V^(mu)delta a delta b.]:}\begin{align*}
\delta V^{\alpha} & =\binom{\text { Change in } V^{\alpha} \text { transported along } \delta a \boldsymbol{e}_{1},}{\text { then } \delta b \boldsymbol{e}_{2}, \text { then }-\delta a \boldsymbol{e}_{1}, \text { then }-\delta b \boldsymbol{e}_{2}} \\
& =\binom{\text { Measure of the curvature }}{\text { of spacetime }}_{\mu} V^{\mu} \delta a \delta b . \tag{11.9}
\end{align*}
This measure of curvature is, once again, supplied by the components of the Riemann tensor R(,,\boldsymbol{R}(,,,).Howdoesthiswork?Inthepreced-) . How does this work? In the preced- ing equation, we have made the deceptively simple, but mathematically significant, claim that the right-hand side of the equation is linear in the sides of the parallelogram delta a\delta a and delta b\delta b. It is this that enables us to identify the part in brackets as the components of a tensor (i.e. a linear slot-machine object). To show that this is the case, we will dive ahead and use this method to derive an expression for the curvature tensor.
Example 11.3
We start with the vector at position AA, with coordinates (a,b)(a, b) and parallel transport it along e_(1)\boldsymbol{e}_{1} to position BB. Parallel transport along e_(1)\boldsymbol{e}_{1} implies grad_(e_(1))V=0\boldsymbol{\nabla}_{\boldsymbol{e}_{1}} \boldsymbol{V}=0, so we must have
The change in the vector V\boldsymbol{V} can be summarized as
{:[(11.16)deltaV^(alpha)=((" Change in "V^(alpha)" transported along "delta ae_(nu),)/(" then "delta be_(mu)," then "-delta ae_(nu)," then "-delta be_(mu)))],[=[(delGamma_(nu beta)^(alpha))/(delx^(mu))-(delGamma^(alpha)_(mu beta))/(delx^(nu))+Gamma_(mu sigma)^(alpha)Gamma_(nu beta)^(sigma)-Gamma_(nu sigma)^(alpha)Gamma_(mu beta)^(sigma)]V^(beta)delta a delta b.]:}\begin{align*}
\delta V^{\alpha} & =\binom{\text { Change in } V^{\alpha} \text { transported along } \delta a \boldsymbol{e}_{\nu},}{\text { then } \delta b \boldsymbol{e}_{\mu}, \text { then }-\delta a \boldsymbol{e}_{\nu}, \text { then }-\delta b \boldsymbol{e}_{\mu}} \tag{11.16}\\
& =\left[\frac{\partial \Gamma_{\nu \beta}^{\alpha}}{\partial x^{\mu}}-\frac{\partial \Gamma^{\alpha}{ }_{\mu \beta}}{\partial x^{\nu}}+\Gamma_{\mu \sigma}^{\alpha} \Gamma_{\nu \beta}^{\sigma}-\Gamma_{\nu \sigma}^{\alpha} \Gamma_{\mu \beta}^{\sigma}\right] V^{\beta} \delta a \delta b .
\end{align*}
As claimed, the part in square brackets depends linearly on the sides of the parallelogram delta a\delta a and delta b\delta b. This enables us to define the Riemann curvature tensor as the (1,3) tensor R(,:}\boldsymbol{R}\left(,\right., , with components ^(11){ }^{11}
Recall for the weak gravitational field metric ds^(2)=-[1+2Phi(x^(i))]dt^(2)+(dx^(2)+dy^(2)+:}\mathrm{d} s^{2}=-\left[1+2 \Phi\left(x^{i}\right)\right] \mathrm{d} t^{2}+\left(\mathrm{d} x^{2}+\mathrm{d} y^{2}+\right.dz^(2)\mathrm{d} z^{2} ) we had non-zero connections
These are identical to the components that we identified earlier in the chapter in eqn 11.4. ^(10){ }^{10} The notation here records the constant coordinate, rather than the limits of the integral. ^(11){ }^{11} Equation 11.19 has the dubious distinction of being one of the more lengthy equations in this subject that it is useful to remember. A helpful aidemémoire that is often employed is to use the matrix form of the equations to say
where square brackets denote a commutator (i.e. [A,B]=AB-BA[A, B]=A B-B A ). If the connection coefficients are written out as matrices, eqn 11.19 has the additional advantage of simplifying the sums over indices. For example, in two dimensions we have
The matrix notation then allows the products of the connection coefficients to be computed using matrix algebra. ^(12){ }^{12} Strictly we should write the final term on the right as (del^(2)Phi)/(delx^(j)delx^(k))g^(ki)\frac{\partial^{2} \Phi}{\partial x^{j} \partial x^{k}} g^{k i} if we were being fussy about tensor components. ^(13){ }^{13} By 'symmetries' here we mean cases where components are, within a sign change, identical to other components. These mean we have to search for fewer distinct components of the tensor to characterize the curvature of spacetime. This is especially important for a (1,3)(1,3) tensor like R\boldsymbol{R}, since is potentially has 4^(4)=2564^{4}=256 different components in (3+1)(3+1)-dimensional spacetime in the absence of symmetry. ^(14){ }^{14} In a general coordinate frame, R\boldsymbol{R} can be computed more directly as follows:
We'll have little need to apply this expression in anger, but it's comforting to know that it exists. Note that the symmetries apply to both frames. ^(15){ }^{15} The symmetries mean that R_(mu mu mu mu)=R_{\mu \mu \mu \mu}=R_(mu mu nu sigma)=R_(mu nu sigma sigma)=0R_{\mu \mu \nu \sigma}=R_{\mu \nu \sigma \sigma}=0.
We stress finally that the curvature tensor is built from connection coefficients and the connection coefficients are build from the metric. The metric field is the source of curvature.
11.4 Symmetries of the Riemann tensor
The Riemann tensor is vitally important object in general relativity. The tensor has a number of important symmetries. ^(13){ }^{13} Let's now find these. We'll consider the all-down index version of R\boldsymbol{R}, achieved by saying R_(alpha beta gamma delta)=g_(alpha mu)R_(beta gamma delta)^(mu)R_{\alpha \beta \gamma \delta}=g_{\alpha \mu} R_{\beta \gamma \delta}^{\mu}.
Example 11.5
The important local flatness theorem tells us it is always possible to find a local inertial frame (LIF) coordinate system such that the metric is identical to the Minkowski metric and the first derivative of the metric field vanishes at a point P\mathcal{P}. We can therefore work at this point P\mathcal{P} where, as a consequence, the connection vanishes such that Gamma^(mu)_(alpha beta)(x=P)=0\Gamma^{\mu}{ }_{\alpha \beta}(x=\mathcal{P})=0. However, the derivatives of the connection do not vanish (since local flatness certainly does not imply the vanishing of the second derivatives of the metric field). Therefore, the Riemann tensor has components given by the derivatives metric field). Therefore, the Riemann tensor has components given by the derivatives
of the connections only. These can, in turn, be expressed as derivatives of the metric of the connections only. These can, in turn, be exp
components, with the result that, in the LIF, ^(14){ }^{14}
This equation, built from the symmetric metric tensor g\boldsymbol{g} makes the symmetries of the curvature tensor R\boldsymbol{R} manifest.
By considering eqn 11.23 we can identify the symmetries of R_(mu nu alpha beta)R_{\mu \nu \alpha \beta}. Swapping indices within the first pair (mu nu)(\mu \nu) and second pair (alpha beta)(\alpha \beta) results in picking up a minus sign, so we have ^(15){ }^{15}
In words, these say that swaps within the first or second pair of indices earn a minus sign; swapping the pairs together does not. We also can check that we have a cyclic identity
One consequence of these symmetries is on the makeup of the Riemann tensor. At first glance, it looks like the R\boldsymbol{R} tensor in NN dimensions has N^(4)N^{4} components. However, these symmetries of the tensor strongly restrict the number of degrees of freedom, so that (as we will prove in Chapter 35) there are only N^(2)(N^(2)-1)//12N^{2}\left(N^{2}-1\right) / 12 independent components. We have for R_(mu nu alpha beta)R_{\mu \nu \alpha \beta} that the number of independent components are as follows.
Some immediate consequences are considered in the following example.
Example 11.6
In one dimension, there is always a coordinate transformation that reduces the metric on a line to the Euclidean form. There is never any curvature. ^(16){ }^{16}
In two dimensions we have a single non-zero component of R\boldsymbol{R}. According to the symmetries, this must be R_(1212)R_{1212}
11.5 The Ricci tensor and Ricci scalar
Once we have the Riemann tensor R\boldsymbol{R} and its symmetries then this also unlocks two simpler tensors that turn out to be essential to building the Einstein equation, linking the curvature and the matter fields of the Universe. It is perhaps to be expected that a (1,3)(1,3) tensor does not feature as raw material in a law of nature. Physics is full of (2,0)(2,0) and (0,2)(0,2) valent tensors, but very few (3,1)(3,1) valent tensors. To build a (0,2)(0,2) tensor from a (3,1)(3,1) tensor the simplest operation would be to contract the up index with a down index. ^(17){ }^{17} The symmetries constrain the possible choices, with the result that we define the Ricci tensor ^(18){ }^{18} as the (0,2)(0,2) valent tensor with components
The symmetries of R\boldsymbol{R} imply that the Ricci tensor is symmetric, such that R_(mu nu)=R_(nu mu)R_{\mu \nu}=R_{\nu \mu}. We can now make a further contraction, and turn the Ricci tensor into another tensor which is exceptionally simple: a (0,0)(0,0) tensor, otherwise known of course as a scalar. We will discuss the resulting object in more detail in the following chapter, but let's define it now. It's called the Ricci scalar RR and is obtained by tracing over the indices of the Ricci tensor. Hence, ^(19){ }^{19} the Ricci scalar RR is given by
The significance of the Ricci tensor will become clear as we try to relate this measure of curvature to the mass-energy of the fields filling the Universe. For now, you can think of it as a kind of average over components of the Riemann tensor. ^(20){ }^{20} ^(16){ }^{16} The one-dimensional example is an interesting, but trivial, case. It has no intrinsic curvature. You can of course embed a one-dimensional space in a higher dimension, and make it appear curved by tying it into a bow or winding it around a cylinder, but that curvature is always extrinsic and never intrinsic. The one-dimensional space is always isometric to a perfectly straight line. Things are more complicated in line. Things are more complicated in
two dimensions; an ant confined to the two dimensions; an ant confined to the
surface of a large ball only knows it is living in a curved space if it goes for a walk around the surface and measures how a vector is parallel transported around its walk and compares the result on returning to its initial position. Such closed-loop walks are not possible in one dimension. ^(17)A{ }^{17} \mathrm{~A} contraction is an operation on a tensor that involves setting an upstairs index equal to a downstairs one and summing. The result is to reduce the number of indices by two. ^(18){ }^{18} Gregorio Ricci-Curbastro (18531925). The Ricci tensor is one of the few important tensors to which we do not assign a bold symbol. The reason is that it does not appear in the field theory of gravity alone, but in a so-called trace-reversed form R_(mu nu)-(1)/(2)g_(mu nu)RR_{\mu \nu}-\frac{1}{2} g_{\mu \nu} R, where RR is the Ricci scalar (the trace over R_(mu nu)R_{\mu \nu}, see below). These are the components of the Einstein tensor G\boldsymbol{G}, which is the quantity given a bold symbol. Some texts do not follow this convention and assign the Ricci tensor a bold symbol such as Ri. ^(19)A{ }^{19} \mathrm{~A} trace is often written as A^(mu)_(mu)=A^{\mu}{ }_{\mu}=A^(0)_(0)+A^(1)_(1)+A^(2)_(2)+A^(3)_(3)A^{0}{ }_{0}+A^{1}{ }_{1}+A^{2}{ }_{2}+A^{3}{ }_{3}. In general relativity, it is best written g_(mu nu)A^(mu nu)g_{\mu \nu} A^{\mu \nu} as an aide-memoir that the metric must be used to construct the trace in curved spacetime. ^(20){ }^{20} It will turn out that it is this average that captures the ability of gravitational curvature to cause volumes to shrink.
11.6 Example computations
We shall now carry out two brute force (and ignorance) calculations of the components of the Riemann tensor, using the equation
Example 11.7
First consider flat space with line element ds^(2)=dr^(2)+r^(2)dtheta^(2)\mathrm{d} s^{2}=\mathrm{d} r^{2}+r^{2} \mathrm{~d} \theta^{2}. We expect that all of the components of the Riemann tensor should vanish. We saw that this metric has connections Gamma^(r)_(theta theta)=-r\Gamma^{r}{ }_{\theta \theta}=-r and Gamma^(theta)_(r theta)=(1)/(r)\Gamma^{\theta}{ }_{r \theta}=\frac{1}{r}. We know from the symmetries that R_(rrr)^(r)=R_(theta theta theta)^(theta)=0R_{r r r}^{r}=R_{\theta \theta \theta}^{\theta}=0. Let's try some of the mixed-index components
^(21){ }^{21} Consequently, the Ricci tensor and Ricci scalar both vanish. This example may not have seemed that exciting, but it's good to check that our formalism works. Flat space is indeed not curved.
As expected, the components all vanish in flat space. ^(21){ }^{21}
Now the (considerably less dull) example of a two-dimensional spherical space on the surface of a sphere of radius aa.
These two examples show how the components of Riemann tensor R\boldsymbol{R} can be evaluated via direct (admittedly, rather tedious) computation. Fortunately, there is a more efficient method to achieve this, although it does rely on a little more geometrical technology, as we'll see in Part V.
11.7 Geodesic deviation revisited
We now have access to the tensor R\boldsymbol{R} that allows us to evaluate the curvature of spacetime. However, as we saw in Chapter 10, our measurements are made in orthonormal frames. By employing a specially-chosen orthonormal frame we can link the components of the Riemann tensor more directly to the notion of geodesic deviation we discussed earlier. The local frame we will choose is the freely falling local inertial frame.
We imagine setting a swarm of particles free from rest and allowing them to follow their geodesics. We assign one the status of being the fiducial geodesic and note its tangent u\boldsymbol{u}. If the particles accelerate relative to each other then we can compute R\boldsymbol{R}. We therefore need to work out the acceleration of a vector xi\boldsymbol{\xi} linking a neighbouring geodesic to the fiducial one, which is related to the Riemann tensor via ^(23){ }^{23}
In the coordinates of a freely falling frame we have ^(24)u^( hat(beta))=(1,0,0,0){ }^{24} u^{\hat{\beta}}=(1,0,0,0), the geodesic deviation equation simplifies to ^(25){ }^{25}
Using the vielbein, we relate the coordinates of R\boldsymbol{R} in the local freely falling frame to a coordinate frame via R^( hat(alpha))_( hat(beta) hat(gamma) hat(delta))=R^{\hat{\alpha}}{ }_{\hat{\beta} \hat{\gamma} \hat{\delta}}=R^(mu)_(nu lambda sigma)(e_(mu))^( hat(alpha))(e_( hat(beta)))^(nu)(e_( hat(gamma)))^(lambda)(e_( hat(delta)))^(sigma)R^{\mu}{ }_{\nu \lambda \sigma}\left(\boldsymbol{e}_{\mu}\right)^{\hat{\alpha}}\left(\boldsymbol{e}_{\hat{\beta}}\right)^{\nu}\left(\boldsymbol{e}_{\hat{\gamma}}\right)^{\lambda}\left(\boldsymbol{e}_{\hat{\delta}}\right)^{\sigma}. We can use the simplified form of the deviation equation in eqn 11.40 to investigate a simple example.
Example 11.9
An important example is the Schwarzschild geometry of a spherically symmetric gravitating object. ^(26){ }^{26} As we shall see, the metric for this spacetime gives rise to a Riemann curvature tensor R\boldsymbol{R} with components in the freely falling frame of
It is also useful to recall that in an orthonormal frame we take u=e_( hat(i))\boldsymbol{u}=\boldsymbol{e}_{\hat{i}}. ^(24){ }^{24} That is, the observer is at rest and so the only non-zero component of u\boldsymbol{u} is u^( hat(t))u^{\hat{t}}. Since in the local coordinate system the observer uses the Minkowski tensor eta\boldsymbol{\eta} to form products, the normalization u*u=-1\boldsymbol{u} \cdot \boldsymbol{u}=-1 becomes eta_( hat(t) hat(t))(u^( hat(t)))^(2)=-1\eta_{\hat{t} \hat{t}}\left(u^{\hat{t}}\right)^{2}=-1 and so, since eta_( hat(t) hat(t))=-1\eta_{\hat{t} \hat{t}}=-1, we have u^( hat(t))=1u^{\hat{t}}=1. ^(25){ }^{25} The use of the freely falling frame allows the swap (D^(2)xi//dtau^(2))^(alpha)\left(\mathrm{D}^{2} \boldsymbol{\xi} / \mathrm{d} \tau^{2}\right)^{\alpha} to d^(2)xi^(alpha)//dtau^(2)\mathrm{d}^{2} \xi^{\alpha} / \mathrm{d} \tau^{2}. Notice how the resulting expression is analogous to a simple harmonic oscillator: x^(¨)=-omega_(0)^(2)x\ddot{x}=-\omega_{0}^{2} x (and it is identical to this equation of motion for hat(alpha)= hat(beta)\hat{\alpha}=\hat{\beta} ). The analogue of the natural frequency of the oscillator is provided by the components of the Riemann tensor. In this way, the curvature of spacetime can be thought of as analogous to a spring constant. ^(26){ }^{26} Discussed in Part IV.
Notice the resemblance of these equations of motion to those of the Newtonian problem discussed in Example 11.1.
Chapter summary
Geodesic deviation allows us to identify tidal forces that result from the curvature of spacetime due to gravity.
The Riemann tensor R\boldsymbol{R} is non-zero if there is intrinsic curvature in spacetime.
The tensor R\boldsymbol{R} can be derived by parallel transporting a vector around a loop. The components of R\boldsymbol{R} are given by
In terms of these variables, the components of R\boldsymbol{R} from the last problem can be written as
{:[R_(rtr)^(t)=-(2M)/(r^(2)(2M-r))","],[R_(theta t theta)^(t)=-(M)/(r)],[R_(phi t phi)^(t)=-(Msin^(2)theta)/(r)],[R_(theta r theta)^(r)=-(M)/(r)],[R_(phi r phi)^(r)=-(Msin^(2)theta)/(r)],[(11.46)R_(phi theta phi)^(theta)=(2Msin^(2)theta)/(r)]:}\begin{align*}
R_{r t r}^{t} & =-\frac{2 M}{r^{2}(2 M-r)}, \\
R_{\theta t \theta}^{t} & =-\frac{M}{r} \\
R_{\phi t \phi}^{t} & =-\frac{M \sin ^{2} \theta}{r} \\
R_{\theta r \theta}^{r} & =-\frac{M}{r} \\
R_{\phi r \phi}^{r} & =-\frac{M \sin ^{2} \theta}{r} \\
R_{\phi \theta \phi}^{\theta} & =\frac{2 M \sin ^{2} \theta}{r} \tag{11.46}
\end{align*}
(a) Write expressions for (i) R_(rrt)^(t)R_{r r t}^{t} and (ii) R_(trt)^(r)R_{t r t}^{r}.
(b) Compute the components of R\boldsymbol{R} in the orthonormal frame.
(11.3) Using the components in Exercise 11.2, show that the components of the Ricci tensor vanish for the Schwarzschild metric outside a star. Why must this be the case?
(11.4) Demonstrate the spacetime described by the rotating frame metric in Exercise 3.5 is flat.
The energy-momentum tensor
The curvature that we discussed in the last chapter provides the lefthand, geometrical side of Einstein's field equation. The right-hand, physical side is supplied by the mass-energy content of the matter fields that fill time and space. ^(1){ }^{1} The mass-energy content of spacetime is expressed using the energy-momentum tensor field T(x)\boldsymbol{T}(x). We met an example of the energy-momentum tensor briefly in Part I of the book. In this chapter, we meet the tensor again in a little more detail, along with its all-important constraint grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0. These tools will enable us, in the next chapter, to put everything together and write down the Einstein field equation for general relativity.
12.1 Another look at the energy-momentum tensor
The energy-momentum tensor field T(x)\boldsymbol{T}(x) gives us access to a tensor T(\boldsymbol{T}(, at each point in spacetime that can be used to evaluate the energy and momentum content of matter fields at that point. If in doubt, you can think of T(x)\boldsymbol{T}(x) as describing the current of momentum for a distribution of mass-energy. We can equally well consider the tensor in (2,0)(2,0) or (0,2)(0,2) form. We start by considering the (0,2)(0,2) version of the tensor
where, since the tensor is symmetric, T_(mu nu)(x)=T_(nu mu)(x).^(2)T_{\mu \nu}(x)=T_{\nu \mu}(x) .{ }^{2}
Take the 4 -velocity of an observer to be u\boldsymbol{u}. There is a recipe for extracting physical information by filling in slots of the tensor T(\boldsymbol{T}(,).) . Start by inserting the velocity into one slot (it doesn't matter which, since the tensor is symmetric), to obtain a 1 -form
{:(12.4)T(","u)=-((" 4-momentum density in ")/(" the observer's rest frame ")).:}\begin{equation*}
\boldsymbol{T}(, \boldsymbol{u})=-\binom{\text { 4-momentum density in }}{\text { the observer's rest frame }} . \tag{12.4}
\end{equation*}
Now insert a unit displacement vector a\boldsymbol{a} into the second slot. Since the slots are full, the output is a number. In this case, we obtain
{:(12.5)T(a","u)=-((" 4-momentum density component directed ")/(" along "a" in the observer's rest frame ")):}\begin{equation*}
\boldsymbol{T}(\boldsymbol{a}, \boldsymbol{u})=-\binom{\text { 4-momentum density component directed }}{\text { along } \boldsymbol{a} \text { in the observer's rest frame }} \tag{12.5}
\end{equation*}
12.1 Another look at the energymomentum tensor 131
12.2 Example energy-momentum tensors
12.3 Classical particles 134
12.4 Conservation laws 136
Chapter summary 139
Exercises 140 ^(1){ }^{1} We expect the Einstein field equation
to read (in words)
Curvature of == Energy density
spacetime == of matter fields
(12.1)
We will (finally!) justify this in the next chapter, using the tools we have developed. ^(2){ }^{2} We saw in Chapter 4, how the (0,2)(0,2) tensor T\boldsymbol{T} could be understood by describing it as an outer product of 1 forms
where tilde(p)\tilde{\boldsymbol{p}} is a momentum 1 -form and tilde(J)\tilde{J} is a number current 1 -form for particles. We previously took tilde(J)\tilde{\boldsymbol{J}} to describe a swarm of non-interacting particles. Although the tensor T\boldsymbol{T} describes energy-momentum more generally than for the case of swarms of particles, this description is useful in understanding T\boldsymbol{T}. For example, if an observer has velocity u\boldsymbol{u}, then we can interpret T(,u)= tilde(p)ox J(u)=- tilde(p)n\boldsymbol{T}(, \boldsymbol{u})=\tilde{\boldsymbol{p}} \otimes \boldsymbol{J}(\boldsymbol{u})=-\tilde{\boldsymbol{p}} n as (minus) the momentum density of particles in the observer's frame, We would inter he T(u)=, tilde(p)(u)ox tilde(J)=- tilde(J)\boldsymbol{T}(\boldsymbol{u})=,\tilde{p}(\boldsymbol{u}) \otimes \tilde{J}=-\tilde{J} pret T(u)=,p(u)ox J=-EJ\boldsymbol{T}(\boldsymbol{u})=,\boldsymbol{p}(\boldsymbol{u}) \otimes \boldsymbol{J}=-E \boldsymbol{J} as (mi(\mathrm{mi} nus) the energy flux. Since the tensor is symmetric, these are the same thing. ^(3){ }^{3} Since we use the local basis vectors in the observer's orthonormal frame, denoted with hats on the indices, the interpretations here will always apply to measurements made in this observer's measurements made in this observer's frame. We can turn these into up components using the Minkowski tensor eta^( hat(mu) hat(nu))\eta^{\hat{\mu} \hat{\nu}}, so we pick up a factor of -1 for each of the two timelike components we raise. This gives us
Fig. 12.1 Flux of the hat(i)\hat{i} th component of momentum across a surface parallel to omega^(j)\boldsymbol{\omega}^{j} (that is, a surface with normal e_(j)\boldsymbol{e}_{j} ).
Filling, instead, both slots with u\boldsymbol{u}, we obtain
{:(12.6)T(u","u)=((" Energy density ")/(" in the observer's rest frame ")):}\begin{equation*}
\boldsymbol{T}(\boldsymbol{u}, \boldsymbol{u})=\binom{\text { Energy density }}{\text { in the observer's rest frame }} \tag{12.6}
\end{equation*}
Example 12.1
We shall also discuss components T_(mu nu)=T(e_(mu),e_(nu))T_{\mu \nu}=\boldsymbol{T}\left(\boldsymbol{e}_{\mu}, \boldsymbol{e}_{\nu}\right). Timelike components can be extracted by noting that in the observer's orthonormal frame u=e_( hat(0))\boldsymbol{u}=\boldsymbol{e}_{\hat{0}} and so, and perhaps most importantly, ^(3){ }^{3}
{:(12.8)T(e_( hat(0)),e_( hat(0)))=T_( hat(0) hat(0))=T^( hat(0) hat(0))=(" energy density ").:}\begin{equation*}
\boldsymbol{T}\left(\boldsymbol{e}_{\hat{0}}, \boldsymbol{e}_{\hat{0}}\right)=T_{\hat{0} \hat{0}}=T^{\hat{0} \hat{0}}=(\text { energy density }) . \tag{12.8}
\end{equation*}
We also have
{:(12.9)T(e_( hat(i)),e_( hat(0)))=T_( hat(i) hat(0))=-T^(i hat(0) hat(0))=-((i" th component of ")/(4"-momentum density ")).:}\begin{equation*}
\boldsymbol{T}\left(\boldsymbol{e}_{\hat{i}}, \boldsymbol{e}_{\hat{0}}\right)=T_{\hat{i} \hat{0}}=-T^{i \hat{0} \hat{0}}=-\binom{i \text { th component of }}{4 \text {-momentum density }} . \tag{12.9}
\end{equation*}
which is also equivalent to an energy flux in the ii th direction. We can also interpret the purely spatial components.
{:(12.10)T(e_( hat(i)),e_( hat(j)))=T_( hat(i) hat(j))=T^( hat(i) hat(j))=((i" th component of 4-momentum flux ")/(" crossing a surface parallel to "omega^( hat(j)))).:}\begin{equation*}
\boldsymbol{T}\left(\boldsymbol{e}_{\hat{i}}, \boldsymbol{e}_{\hat{j}}\right)=T_{\hat{i} \hat{j}}=T^{\hat{i} \hat{j}}=\binom{i \text { th component of 4-momentum flux }}{\text { crossing a surface parallel to } \boldsymbol{\omega}^{\hat{j}}} . \tag{12.10}
\end{equation*}
The idea of the latter is shown in Fig. 12.1. From the version of the energy-momentum tensor in the sidenote, we have that T_( hat(i) hat(j))= tilde(p)(e_( hat(i))) tilde(J)(e_( tilde(j)))=p_( hat(i))J_( hat(j))T_{\hat{i} \hat{j}}=\tilde{p}\left(e_{\hat{i}}\right) \tilde{J}\left(e_{\tilde{j}}\right)=p_{\hat{i}} J_{\hat{j}}, which is consistent with this interpretation. However, another interpretation is possible. Since rate of momentum is equivalent to force, we can also write
{:(12.11)T(e_( hat(i)),e_( hat(j)))=T_( hat(i) hat(j))=T^( hat(i) hat(j))=((i" th component of force ")/(" across a surface parallel to "omega^( hat(j)))).:}\begin{equation*}
\boldsymbol{T}\left(\boldsymbol{e}_{\hat{i}}, \boldsymbol{e}_{\hat{j}}\right)=T_{\hat{i} \hat{j}}=T^{\hat{i} \hat{j}}=\binom{i \text { th component of force }}{\text { across a surface parallel to } \boldsymbol{\omega}^{\hat{j}}} . \tag{12.11}
\end{equation*}
A cartoon of the components of the energy-momentum tensor is shown in Fig. 12.2.
Fig. 12.2 The components of the energy-momentum tensor in the orthonormal frame of an observer.
12.2 Example energy-momentum tensors
The general discussion of the energy-momentum tensor is all very well, but of limited use without some idea of what the various components of T\boldsymbol{T} actually are. We now examine some examples of how to build ^(4)T{ }^{4} \boldsymbol{T}. Since gravity works over long length scales where the granularity of matter is often not too important, we will mostly be concerned with
continuous fields of matter and so we need the energy-momentum tensor for this purpose. The continuous version of the particle matter we have considered so far is a fluid. We will consider a so-called perfect fluid, which is continuous matter with energy density rho\rho and an isotropic pressure ^(5)p{ }^{5} p in its rest frame, but no other interactions. ^(6){ }^{6}
Example 12.2
First, consider the case of the fluid at rest in a local inertial reference frame, which has metric eta\boldsymbol{\eta}. We have an energy density T^( hat(0) hat(0))=T_( hat(0) hat(0))=rhoT^{\hat{0} \hat{0}}=T_{\hat{0} \hat{0}}=\rho, where rho\rho is the total energy-density of the fluid, as measured in the fluid's rest frame. The force in the hat(i)\hat{i} th direction, perpendicular to the surface whose normal is in the hat(i)\hat{i} direction, is simply the definition of the pressure pp. We therefore have T^(ii)=T_( hat(i) hat(i))=pT^{i i}=T_{\hat{i} \hat{i}}=p and can write
The velocity 4 -vector u\boldsymbol{u} of the fluid fixes the timelike basis vector according to e_( hat(0))=\boldsymbol{e}_{\hat{0}}= u\boldsymbol{u}. In the fluid's rest frame, the velocity 1 -form tilde(u)=u_( hat(mu))omega^(mu)\tilde{\boldsymbol{u}}=u_{\hat{\mu}} \boldsymbol{\omega}^{\mu} has components u_( hat(mu))=u_{\hat{\mu}}=(-1,0,0,0)(-1,0,0,0). The ( 0,2 ) tensor product tilde(u)ox tilde(u)\tilde{\boldsymbol{u}} \otimes \tilde{\boldsymbol{u}} has components
The components of this equation balance ^(7){ }^{7} and so it represents a valid tensor equation.
We conclude that for a perfect fluid in a local inertial frame (or in flat spacetime) we have a equation for the (0,2)(0,2) tensor T\boldsymbol{T}, which we can write as
where we take tilde(u)\tilde{\boldsymbol{u}} to be the velocity 1 -form of the fluid itself. The tensor T\boldsymbol{T} inputs two vectors to generate a number. It has components
Thermodynamics says that a photon gas has the equation of state rho=3p\rho=3 p, so we conclude that the energy-momentum tensor for the photon gas is traceless (T=0).^(8)(T=0) .{ }^{8} ^(5){ }^{5} Do not confuse pressure pp with momentum, which regrettably has the same symbol. ^(6){ }^{6} The total energy density rho\rho will generally include a contribution from the mass density rho_(0)\rho_{0} of the fluid, along with a contribution from the internal energy density. The latter results from internal motion and interactions between the constituents of the fluid. If there is no internal energy then pressure p=0p=0 and we call the matter dust, which has rho=rho_(0)\rho=\rho_{0}. In contrast, a perfect fluid has an isotropic pressure pp in its rest frame, but no viscosity or shear stresses (and now rho!=rho_(0)\rho \neq \rho_{0} ). The relationship between rho\rho and pp for the fluid is known as its equation of state. ^(7){ }^{7} That is, we have hat(mu) hat(nu)\hat{\mu} \hat{\nu} in the down position on both sides. Recall that we say such equations are manifestly covariant and are therefore 'valid tensor equations'. ^(8){ }^{8} A field with an energy-momentum tensor whose trace vanishes is expected to have massless excitations. This is true for the electromagnetic field, whose excitations, photons, are certainly massless. The vanishing trace can be linked to scale invariance, as discussed in the book by Zee.
The equivalence principle tells us that a valid tensor equation in a flat space is a valid tensor equation in a curved space as long as we replace the Minkowski metric tensor eta\boldsymbol{\eta} with the generalized metric tensor g\boldsymbol{g}. We can then upgrade eqn 12.16 to the generalized version in curved space, where it becomes
Since tensor equations in physics are generally valid with respect to transforming all indices from up to down, and vice versa, we shall also use the ( 2,0 ) version of T\boldsymbol{T} whose components in a general coordinate frame are T^(mu nu)=(rho+p)u^(mu)u^(nu)+pg^(mu nu)T^{\mu \nu}=(\rho+p) u^{\mu} u^{\nu}+p g^{\mu \nu}. In comparing quantities with up and down indices, remember that, as usual, indices are raised and lowered using the components of the metric tensor.
12.3 Classical particles
Einstein's equation is expressed in terms of classical fields. We shall find that the matter field corresponding to a perfect fluid is the most useful for our applications. However, in addition to fluids, we can also treat the motion of point-like particles. The most simple example is introduced below.
Example 12.4
Consider a swarm of nn (non-interacting) dust particles per unit volume in flat space, each of rest mass mm, moving with the same 4 -velocity with components (gamma,gammav^(x),gammav^(y),0)\left(\gamma, \gamma v^{x}, \gamma v^{y}, 0\right). Temporarily restoring factors of cc, we have total energy density (gamma,gammav^(x),gammav^(y),0)\left(\gamma, \gamma v^{x}, \gamma v^{y}, 0\right). Temporarily restoring factors of cc, we have total energy density T^(00)=nm gammac^(2)T^{00}=n m \gamma c^{2}. The momentum density in the ii-direction is T^(0i)=T^(i0)=nm gamma cv^(i)T^{0 i}=T^{i 0}=n m \gamma c v^{i}. The momentum flux in, say, the xx-direction, crossing the surface parallel to omega^(y)\boldsymbol{\omega}^{y} is T^(xy)=nm gammav^(x)v^(y)T^{x y}=n m \gamma v^{x} v^{y}. This is enough for us to construct the tensor, whose components are
{:(12.20)T^(mu nu)=nm gamma([c^(2),v^(x)c,v^(y)c,0],[v^(x)c,(v^(x))^(2),v^(x)v^(y),0],[v^(y)c,v^(x)v^(y),(v^(y))^(2),0],[0,0,0,0]):}T^{\mu \nu}=n m \gamma\left(\begin{array}{cccc}
c^{2} & v^{x} c & v^{y} c & 0 \tag{12.20}\\
v^{x} c & \left(v^{x}\right)^{2} & v^{x} v^{y} & 0 \\
v^{y} c & v^{x} v^{y} & \left(v^{y}\right)^{2} & 0 \\
0 & 0 & 0 & 0
\end{array}\right)
In order to treat particles more generally using the techniques of field theory, we can use the fact that particles appear on a world line z^(mu)(tau)z^{\mu}(\tau), where tau\tau is the proper time parametrizing the world line. ^(9){ }^{9} In order to select that part of spacetime that the particle intersects, we use a delta function to pick out the curve z^(mu)(tau)z^{\mu}(\tau). The particles are assumed noninteracting here, so the only contribution to their energy is their mass. To capture the mass density field rho_(0)(x)\rho_{0}(x) of a particle of mass mm in flat spacetime we write
To see how the definition works, we change the integration variable from tau\tau to z^(0)(tau)z^{0}(\tau) using u^(0)=dz^(0)//dtauu^{0}=\mathrm{d} z^{0} / \mathrm{d} \tau to write
where tau\tau in the final line solves the equation z^(0)(tau)=tz^{0}(\tau)=t. This is now simply a function that finds the particle in three-dimensional space. Notice how the factor u^(0)u^{0}, which in flat space is simply gamma\gamma, deals with the length contraction of the three-dimensional volume that determines the value of the mass density (see Fig. 2.7).
We can also write the mass current, which is most generally written as a field J_(m)(x)=rho_(0)(x)u(x)\boldsymbol{J}_{m}(x)=\rho_{0}(x) \boldsymbol{u}(x), where u(x)\boldsymbol{u}(x) is the velocity of the matter at xx. In terms of particles, we have the component expression ^(10){ }^{10}
which is simply a version of eqn 12.19 for particles with p=0p=0.
Example 12.6
In curved spacetime, we must recall that sqrt(-g)d^(4)x\sqrt{-g} \mathrm{~d}^{4} x is the invariant volume element, so delta^((4))(x-y)//sqrt(-g)\delta^{(4)}(x-y) / \sqrt{-g} is a scalar. The expressions above become, in curved spacetime,
^(11){ }^{11} Restoring factors of cc we have the 4 -vector J^(mu)=(rho c, vec(J))J^{\mu}=(\rho c, \vec{J}). The spacelike part vec(J)\vec{J} therefore has units of charge density times velocity. ^(12){ }^{12} Integrating eqn 12.30 over a flatspace 3 -volume and invoking the divergence theorem gives the conserva tion law in three-dimensional Cartesian space as
where SS is the boundary of the volume VV, whose area element is written as d vec(S)\mathrm{d} \vec{S} In words, the rate of change of charge in the volume is equal to the flux of particles through its surface. ^(13){ }^{13} The 3-surface could be oriented, so that we have a sense of direction de fined from one side of the surface to the other, just as one has for the 2-surface in three-dimensions. ^(14){ }^{14} This can be interpreted as saying that all the charge that enters the vol ume in the past part of its boundary (where J\boldsymbol{J} and dSigma\mathrm{d} \boldsymbol{\Sigma} might point in opposite directions) has to exit out of the future part of its boundary (where J\boldsymbol{J} and dSigma\mathrm{d} \boldsymbol{\Sigma} might point in the same direction).
Fig. 12.3 A volume V\mathcal{V} in flat spacetime is bounded by a 3 -surface delV\partial \mathcal{V} which can be described by lots of infinitesimal surface vectors dSigma\mathrm{d} \boldsymbol{\Sigma}. The current density 4-4- vector J\boldsymbol{J} flows into and out of the surface, subject to a global conservation law (eqn 12.32).
Example 12.7
In general, a current vector J\boldsymbol{J} has components J^(mu)=(rho, vec(J))J^{\mu}=(\rho, \vec{J}). The timelike component rho\rho is called a charge density. In flat space, the integral of this quantity over the 3 -volume VV is the conserved charge
{:(12.28)Q=int_(V)d^(3)x rho:}\begin{equation*}
Q=\int_{V} \mathrm{~d}^{3} x \rho \tag{12.28}
\end{equation*}
This charge might be electrical charge, or might be the number density of particles. The spacelike components represents the flux of the charge, that is, the number of charges that cross unit area per unit time. ^(11){ }^{11} Again working in flat spacetime, local conservation implies that J\boldsymbol{J} obeys the equation ^(12){ }^{12}
{:(12.31)del_(mu)J^(mu)=0","quad" or equivalently, "quadJ^(mu)_(,mu)=0:}\begin{equation*}
\partial_{\mu} J^{\mu}=0, \quad \text { or equivalently, } \quad J^{\mu}{ }_{, \mu}=0 \tag{12.31}
\end{equation*}
We shall often write this in vector notation as grad*J=0\boldsymbol{\nabla} \cdot \boldsymbol{J}=0 (i.e. the divergence of J\boldsymbol{J} is zero). This expression guarantees that the rate of change of charge is equal to the amount of the charge flowing into, or out of, an element of volume. We call this local conservation of charge. We can't therefore have charge disappear on one point and then appear at some other arbitrary point in the Universe, which is what we mean by 'local' in our definition. (Conversely, if we wanted to guarantee conservation of charge, but allow charges to arbitrarily transport across the universe, we would demand only global conservation of a charge.)
We can think about this problem another way. Imagine a flat 3 -surface which lies in the xyt plane, perpendicular to the zz-axis. Then J^(z)J^{z} represents the total charge that flows across this surface per unit area, per unit time. If this 3 -surface had dimensions Sigma_(z)=a xx b xx tau\Sigma_{z}=a \times b \times \tau (i.e. a xx ba \times b in the xyx y plane and tau\tau in the time dimension) then J^(z)Sigma_(z)J^{z} \Sigma_{z} would represent the total charge flowing through this 3 -surface (we've multiplied through by Sigma_(z)\Sigma_{z} so it is no longer 'per unit area, per unit time'). We could generalize this idea (which has privileged a particular slice in space time) and think of J^(alpha)Sigma_(alpha)J^{\alpha} \Sigma_{\alpha} as the total charge that flows across any flat 3 -surface described by a 4 -vector Sigma\boldsymbol{\Sigma} normal to the 3-surface. ^(13){ }^{13} We could make this even more general by allowing the 3 -surface to be curved, and then the total charge that flows across it could be written as intJ^(mu)dSigma_(mu)\int J^{\mu} \mathrm{d} \Sigma_{\mu}; here, both J\boldsymbol{J} and Sigma\boldsymbol{\Sigma} are now functions of spacetime.
Now consider some 4 -volume of spacetime V\mathcal{V}, bounded by a closed 3 -surface which we can write as delV\partial \mathcal{V} (see Fig. 12.3). Integrating over the closed surface delV\partial \mathcal{V}, we can then express the global law of charge conservation as ^(14){ }^{14}
This expresses global charge conservation in flat spacetime, since our 4 -volume V\mathcal{V} can be of any size.
In setting this problem up, we have been thinking about the conservation of electric charge, but this approach would work with any conserved scalar quantity. For example, it could be a number-flux of baryons (which one could write as N=(N^(0),( vec(N)))N=\left(N^{0}, \vec{N}\right), with N_(0)N_{0} the baryon number density and NN the baryon number 3-flux) or the flux of rest mass [rho_(0)(1,( vec(u))):}\left[\rho_{0}(1, \vec{u})\right., with rho_(0)\rho_{0} the rest mass density and vec(u)\vec{u} the 3 -velocity]. For both these cases, an equation analogous to eqn 12.32 would follow.
Energy and momentum are conserved, so we might expect a local conservation law for the energy-momentum tensor (which describes the flow of 4 -momentum), just as we found for J\boldsymbol{J} (which describes the flow of
charge). By analogy with the previous example, the local conservation law in flat spacetime can be written in three (equivalent) ways
Putting nu=0\nu=0 into T^(mu nu)_(,mu)=0T^{\mu \nu}{ }_{, \mu}=0 gives T^(00)_(,0)+T^(0i)_(,i)=0T^{00}{ }_{, 0}+T^{0 i}{ }_{, i}=0, which expands into del rho//del t+ vec(grad)*(rho vec(v))=0\partial \rho / \partial t+\vec{\nabla} \cdot(\rho \vec{v})=0, which is the conservation law given in eqn 12.30 . This part therefore expresses local conservation of energy in flat spacetime.
Putting nu=i\nu=i into T^(mu nu)_(,mu)=0T^{\mu \nu}{ }_{, \mu}=0 gives T^(i0)_(,0)+T^(ij)_(,j)=0T^{i 0}{ }_{, 0}+T^{i j}{ }_{, j}=0, which expands into del(rhov^(i))//del t+ vec(grad)*(rhov^(i)( vec(v)))=0\partial\left(\rho v^{i}\right) / \partial t+\vec{\nabla} \cdot\left(\rho v^{i} \vec{v}\right)=0, which is a very similar conservation law, but now the ii th component of the momentum density rhov^(i)\rho v^{i} has replaced the energy. This part therefore expresses local conservation of momentum in flat spacetime.
All this has been for flat spacetime. The next natural step is check if a version of T^(mu nu)_(,mu)=0T^{\mu \nu}{ }_{, \mu}=0 is valid for curved spacetime. If an observer makes measurements in a local inertial frame (LIF), they will conclude that
Recall that, in a LIF, the connection coefficients Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta} vanish at a point, ^(15){ }^{15} making ordinary and covariant derivatives identical. As a result, the observer will conclude that
This is a valid tensor equation in flat space so, by the principle of general covariance, should also work in curved space, if properly upgraded. The upgrade simply involves removing the hats
and we now have our curved space version. This argument, based on the principle of general covariance, is often called comma goes to semicolon and is equivalent to the swap eta rarr g\boldsymbol{\eta} \rightarrow \boldsymbol{g}. Probably the most memorable way to write this important result, eqn 12.36 , is ^(16){ }^{16}
Example 12.9
We can take the divergence of the perfect fluid tensor part by part by noting that the covariant derivative obeys the Leibniz ( -=\equiv chain) rule. Start with components T^(mu nu)=(rho+p)u^(mu)u^(nu)+pg^(mu nu)T^{\mu \nu}=(\rho+p) u^{\mu} u^{\nu}+p g^{\mu \nu} and using the semicolon notation and the Leibniz rule, we find ^(17){ }^{17}
^(15){ }^{15} This argument is the same one we rehearsed in Example 8.8. ^(16){ }^{16} We have not yet worried about how to take the covariant derivative of anything more complicated than a vector. We return to this in Part V, but for now note that, written explicitly, the component equation is
^(17){ }^{17} Remember that the tensor T\boldsymbol{T} is symmetric, so T^(mu nu)=T^(nu mu)T^{\mu \nu}=T^{\nu \mu} and there is no metric, so T^(mu nu)=T^(nu mu)T^{\mu \nu}=T^{\nu \mu} and there is no
difference if we take the derivative with respect to mu\mu or nu\nu. Note also that (i) for a scalar function ff we have
f_(;mu)=f_(,mu)f_{; \mu}=f_{, \mu}
and (ii) we have the important equation
g^(mu nu)_(;mu)=0g^{\mu \nu}{ }_{; \mu}=0
^(18){ }^{18} This can be reduced to the geodesi equation in the case of constant pres sure and conserved mass density. See the exercises at the end of this chapter and also Chapter 39. ^(19){ }^{19} As always, the geometrical interpre tation of velocity is the vector tangent to the world line of a observer.
Fig. 12.4 The world lines of three parallel observers in flat spacetime. With respect to the coordinate frame of the figure, they all have identical velocity.
This quantity must, of course, vanish since grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0. Noting the comments in the sidenote, we find
We shall see in Chapter 39 that this complicated expression gives us an equation of motion for the fluid. ^(18){ }^{18}
We therefore have that grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0 is a valid constraint on the energymomentum tensor T\boldsymbol{T} in curved space. It's tempting to interpret grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0 as an equation of energy conservation for curved spacetime, but we need to be careful. The conservation of energy over anything other than small distances in curved spacetime is a knotty problem.
To see this, let's first discuss what we mean by the energy EE in general relativity. We define the energy of an object with momentum 1 -form tilde(p)\tilde{\boldsymbol{p}} via E=- tilde(p)(u)E=-\tilde{\boldsymbol{p}}(\boldsymbol{u}), where u\boldsymbol{u} is the velocity ^(19){ }^{19} of an observer who is located at the site of the object whose energy we are measuring. In flat space, we then define the energy measured by an observer who isn't present, as being equal to the energy measured by the local observer if their velocity is parallel to that of the distant observer. Geometrically, parallel observers have no relative velocity in flat spacetime, so their world lines are parallel straight lines (Fig. 12.4). We can define a velocity field u(x)\boldsymbol{u}(x) for all observers, which when we input a position in spacetime, will output a vector tangent to the world line of whichever observer is at that spacetime point. Mathematically, the condition that these flat-spacetime observers have parallel velocity may be written as
which tells us that the tangent vectors to the world lines don't vary in space, and so are parallel. The mathematical need for parallel observers defined in this way is justified in the next example.
Example 12.10
In flat spacetime, the vanishing divergence is written as T^(mu nu),mu=0T^{\mu \nu}, \mu=0. Recall that for dust we had T= tilde(J)ox tilde(p)\boldsymbol{T}=\tilde{\boldsymbol{J}} \otimes \tilde{\boldsymbol{p}} and so the energy density can be written as
The quantity tilde(M)()=E tilde(J)()\tilde{\boldsymbol{M}}()=E \tilde{\boldsymbol{J}}() can be thought of as the current 1-form of energy carried by the particles in the dust cloud. If this is to be locally conserved, we must hav M^(mu)_(,mu)=0M^{\mu}{ }_{, \mu}=0, where M^(mu)M^{\mu} are the components of vector M\boldsymbol{M} which, from eqn 12.43 are
where we have used the Leibniz rule and T^(mu nu)_(,mu)=0T^{\mu \nu}{ }_{, \mu}=0. To guarantee that this vanishes, we require u^(nu)_(,mu)=0u^{\nu}{ }_{, \mu}=0. Physically, this requires us to have access to a family of inertial, parallel observers if we are to have a notion of conservation of energy.
The argument above is appropriate for flat spacetime. However, in curved spacetime we lose the ability to globally define a family of parallel observers, and this will frustrate our attempts to interpret grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0 as statement of energy conservation. This is because 'being parallel' is a path-dependent property in curved spacetime, as we see in the next example. ^(20){ }^{20}
Example 12.11
Replaying the last example for curved spacetime, we apply a condition grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0 using the covariant derivative, and so we need
Follow the same steps as before, we conclude that this is so if u_(nu;mu)=0u_{\nu ; \mu}=0, which, owing to the symmetry of T\boldsymbol{T}, can be rewritten as
This condition is known as Killing's equation (Chapter 33) and is not generally true of velocity vector fields in curved spacetime. ^(21){ }^{21}
The previous example implies that in curved spacetime, although grad*T=\boldsymbol{\nabla} \cdot \boldsymbol{T}= 0 holds, it can't be strictly interpreted in terms of energy conservation. Physically, this is to be expected since general relativity shows that matter and gravity are coupled: the gravitational field (described by the curvature of spacetime) can do work on matter, and matter can do work on the gravitational field. However, over small distances they don't do much work on each other and so, approximately at least, we have energy conservation ^(22){ }^{22} since u_(nu;mu)~~0u_{\nu ; \mu} \approx 0.
We now have the object T\boldsymbol{T} that, in some form, lives on the right-hand side of Einstein's equation, along with its key property ^(23)grad*T=0{ }^{23} \boldsymbol{\nabla} \cdot \boldsymbol{T}=0 (which is true, in spite of its rather subtle interpretation). Our next task is to marry the geometric left-hand side of Einstein's equation with the physical right-hand side.
Chapter summary
The right-hand side of Einstein's equation encodes the energymomentum of matter fields using the tensor T\boldsymbol{T}.
In flat spacetime, the local conservation of energy-momentum can be expressed via the important constraint grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0. The same expression holds in curved spacetime. ^(20){ }^{20} Another way of seeing this point is that, to globally conserve momentum, we need to demonstrate
where delV\partial \mathcal{V} is the boundary of some closed region of spacetime V\mathcal{V}. But to work this out you have to combine lots of individual 4 -vectors T^(mu nu)dSigma_(nu)T^{\mu \nu} \mathrm{d} \Sigma_{\nu} from different parts of the surface and bring them to a common location to add them to a com add them all up. No problem in flat space-
time, but in curved spacetime they all live in different tangent spaces and this live in different tangent spaces and this
process cannot be done in a well-defined manner. ^(21){ }^{21} Special cases that do obey Killing's equation are very interesting and we exequation are very interesting and we ex-
amine them in detail from Part IV onwards. ^(22){ }^{22} We shall therefore call grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0 a conservation equation in this book, in a slight abuse of language. The law grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0 can be proven geometrically from the invariance of the manifold with respect to a very general type of translation known as a diffeomorphism. (See Appendix C for more details.) So although it does not represent conservation of energy in the strict sense, grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0 is still a significant constraint on our equations. ^(23){ }^{23} We also remark that grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0 is connected with the geodesic principle, namely that free massive point particles follow timelike geodesics. The conservation equation captures the idea that the massive particle is free, that is, it is not exchanging energy-momentum with the fields and matter in its environment. Together with an energy condition (see page 145 for more details) to capture the idea that energy propagates within the body in a timelike or null manner, the geodesic principle can be proved (see e.g. J. Ehlers and R. Geroch, Ann. Phys. 309, 232 (2004) for a proof, and J. O. Weatherall, Studies in the History and Philosophy of Modern Physics 42, 276 (2011) for a reflection Physics 42, 276 (2011) for a reflection
on the debate that has ensued about on the debate that has ens
the status of the principle).
Exercises
(12.1) Consider the transformations between Cartesian coordinates (t,x,y,z)(t, x, y, z) and spherical polar coordinates (t,r,theta,phi)(t, r, \theta, \phi). In the Cartesian system, the (0,2)(0,2) tensor T\boldsymbol{T} is diagonal with non-zero components
where i=x,y,zi=x, y, z. Transform the components of this tensor to spherical polars.
(12.2) Consider dust with energy density T\boldsymbol{T} with components T^(mu nu)=rho_(0)u^(mu)u^(nu)T^{\mu \nu}=\rho_{0} u^{\mu} u^{\nu}.
(a) Explain why grad*(rho_(0)u)=0\boldsymbol{\nabla} \cdot\left(\rho_{0} \boldsymbol{u}\right)=0.
(b) Show that
(c) Using the result of part (a) show that the con servation of mass-energy guarantees that the dust particles obey the geodesic equation and hence that dust particles follow geodesic world lines.
(12.3) We write an energy-momentum tensor for a particle as
where z(tau)z(\tau) is the world line of the particle parametrized by proper time tau\tau. A useful expression when using this tensor, is that for functions f(x)f(x) and g(x)g(x) we have
{:(12.51)intdx delta[f(x)]g(x)=sum_(a)(g(x_(a)))/(|f^(')(x_(a))|):}\begin{equation*}
\int \mathrm{d} x \delta[f(x)] g(x)=\sum_{a} \frac{g\left(x_{a}\right)}{\left|f^{\prime}\left(x_{a}\right)\right|} \tag{12.51}
\end{equation*}
where the sum over aa is over all values of x_(a)x_{a} that have the property that f(x_(a))=0f\left(x_{a}\right)=0.
Using the tools above, justify the following expressions for the components of the energy-momentum tensor for a swarm of particles in flat space.
(a) The momentum density
where the index nn labels a specific particle and tau_(n)\tau_{n} solves the equation z^(0)(tau_(n))=tz^{0}\left(\tau_{n}\right)=t and tt is the coordinate time [i.e. particle nn 's proper time at the coordinate time tt that is inputted into the tensor T(x)T(x) as part of the argument xx with components {:x^(mu)=(t,( vec(x)))]\left.x^{\mu}=(t, \vec{x})\right].
(12.4) By applying grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0 to the energy-momentum tensor for a single particle in the previous question, show that we obtain the geodesic equation.
(12.5) In flat space, define a (1,1)(1,1) energy-momentum tensor for dust particles as
This (1,1)(1,1) object has two slots: one for a 1 -form and one for a vector. It has components T^(mu)_(nu)T^{\mu}{ }_{\nu}.
(a) Find the eigenvalue of this tensor when the velocity eigenvector v\boldsymbol{v} is inserted. That is, determine the constants alpha\alpha for the equation
Note that the other three eigenvectors are orthogonal to v\boldsymbol{v} and therefore in the 3-space we can call v^(_|_)\boldsymbol{v}^{\perp} By isotropy these three eigenvectors are degenerate and have some eigenvalue pp, such that
See the book by Ludvigsen for a discussion of this approach.
The gravitational field equations
The key idea of general relativity is that the curvature of spacetime, which gives rise to gravitation, is determined by the energy density of the matter fields of the Universe, such that we can write the Einstein equation
{:(13.1)((" Curvature of ")/(" spacetime "))=((" Energy density ")/(" of matter fields ")).:}\begin{equation*}
\binom{\text { Curvature of }}{\text { spacetime }}=\binom{\text { Energy density }}{\text { of matter fields }} . \tag{13.1}
\end{equation*}
At last, in this chapter, we are able to formulate the details of this equation! Our task is, on the (geometrical) left-hand side, to identify a suitable expression for the curvature of spacetime and, on the (physical) right, a suitable expression for the density of energy in matter fields.
In order to get to Einstein's equation, we take the route shown in Fig. 13.1. Starting with the physical principles of fields and of eqn 13.1, we shall review the key elements of geometry that supply the left-hand side of the equation. We then examine the physics of the right-hand side, in particular the energy-momentum tensor field T(x)\boldsymbol{T}(x) that encodes the energy density of the matter fields that fill the Universe. Finally we link these to form the Einstein field equation, the tensor field version of eqn 13.1.
13.1 Geometry: a recap of the key ingredients
The metric tensor field g(x)\boldsymbol{g}(x) is the foundation of the curvature of spacetime. As we've written it, it is a classical field: a function of position in spacetime xx that gives us a tensor valid at the point xx. We often express the components of the metric field g_(mu nu)(x)g_{\mu \nu}(x) via a line element
There is no way to tell whether a single particle accelerates due to the effects of a homogeneous gravitational field or because of the choice of coordinates. If a gravitational field is present, and all real-life gravitational fields are necessarily inhomogeneous, spacetime is curved and this gives rise to the acceleration of a particle in the absence of forces external to spacetime. However, a particle might accelerate merely due to the perversity of the coordinate system that we've chosen, rather than due to curvature. The method to unambiguously detect curvature
13.1 Geometry: a recap of the key ingredients
Fig. 13.1 A conceptual route to Einstein's equation. The left-hand side leads to the Einstein tensor field G\boldsymbol{G}, related to the Riemann tensor field as described later in this chapter. The right-hand side leads to a constant 8pi G8 \pi G (where GG here is the gravitational constant) multiplied by the energymomentum tensor field T\boldsymbol{T}. ^(1){ }^{1} The connection coefficients may be re lated to the metric tensor field via the component equation
where xi\boldsymbol{\xi} is the particles' separation and u\boldsymbol{u} is the velocity of the particle follow ing the fiducial geodesic. ^(3){ }^{3} Note for later in the chapter that this is symmetric: R_(mu nu)(x)=R_(nu mu)(x)R_{\mu \nu}(x)=R_{\nu \mu}(x). ^(4){ }^{4} Recall that a scalar is the same, independent of the coordinate system in which it is evaluated. ^(5){ }^{5} One point of view is that the (1,3)(1,3) Rie mann curvature tensor field R(x)\boldsymbol{R}(x) is the gravitational field. So although we have the equivalence principle telling us that a measurement at a point cannot tell us whether it's gravitation or accelera tion causing an effect, we can identify the physical field that causes gravitation. We cannot simply slot this 'gravitational field R\boldsymbol{R} into the Einstein equation (curvature) == (energy), owing to the ( 1,3 ) valence of R\boldsymbol{R} compare to the (2,0)(2,0) valence of the energy-momentum (2,0) TT However, if we momentum tensor TT. Hower, if we are looking for the physical manifestation of grav itation, this is as good a candidate a any. ^(6){ }^{6} See also Section 0.4.
involves transforming away the effects due to the coordinate system to arrive at what's left behind: the curvature due to gravitation. This is the motivation behind the Riemann curvature tensor R(x)\boldsymbol{R}(x), which evaluates the presence of this stuff that's left behind. We expect that this tensor should play a key role in the left-hand side of the equation, which indeed it does. It depends on the (first and second) derivatives of the metric field. The first derivatives of g(x)\boldsymbol{g}(x) provide the connection coefficients Gamma^(mu)_(alpha beta)(x),^(1)\Gamma^{\mu}{ }_{\alpha \beta}(x),{ }^{1} and reflect how the metric causes the coordinates to change as we move through spacetime. They are easily manipulated, so we often describe R(x)\boldsymbol{R}(x) in terms of them, rather directly through g(x)\boldsymbol{g}(x).
To obtain R(x)\boldsymbol{R}(x) and thus determine the curvature of spacetime, we evaluate the geodesic deviation of two freely falling particles whose motion was initially parallel and extract R(x)\boldsymbol{R}(x) from the equation of motion of their separation. ^(2){ }^{2} The resulting Riemann tensor field has components R^(alpha)_(beta gamma delta)(x)=(delGamma^(alpha)_(delta beta)(x))/(delx^(gamma))-(delGamma^(alpha)_(gamma beta)(x))/(delx^(delta))+Gamma^(alpha)_(gamma mu)(x)Gamma^(mu)_(delta beta)(x)-Gamma^(alpha)_(delta mu)(x)Gamma^(mu)_(gamma beta)(x)R^{\alpha}{ }_{\beta \gamma \delta}(x)=\frac{\partial \Gamma^{\alpha}{ }_{\delta \beta}(x)}{\partial x^{\gamma}}-\frac{\partial \Gamma^{\alpha}{ }_{\gamma \beta}(x)}{\partial x^{\delta}}+\Gamma^{\alpha}{ }_{\gamma \mu}(x) \Gamma^{\mu}{ }_{\delta \beta}(x)-\Gamma^{\alpha}{ }_{\delta \mu}(x) \Gamma^{\mu}{ }_{\gamma \beta}(x).
This gives us access to the curvature at each point in spacetime. Two quantities derived from the Riemann tensor field will be important in setting up the Einstein equation. The first is the Ricci tensor field, which at a spacetime point xx has components ^(3){ }^{3}
Using these two questions, we will be able to gain access to the curvature of spacetime. This will be achieved by taking a double derivative of the metric field g(x)\boldsymbol{g}(x), which is the basis of the left-hand side of the Einstein equation. ^(5){ }^{5} After having reviewed the geometrical left-hand side of the equation, let's now turn to the right-hand side and the physical content of the theory.
13.2 Physics: the key ingredients
We start with a principle: a physical relativistic/geometrical field theory of gravitation must be compatible with Newton's theory in the nonrelativistic limit. It is therefore useful to revisit Newton's law of gravitation from the point of view of fields and see what it teaches us about the shape that our new theory must take. ^(6){ }^{6} Newton's law says that the force vec(F)\vec{F} experienced by a mass mm from a gravitating object with mass MM separated by a displacement vector vec(r)\vec{r}, is given by
{:(13.8) vec(F)(r)=-(GMm)/(r^(3))* vec(r)",":}\begin{equation*}
\vec{F}(r)=-\frac{G M m}{r^{3}} \cdot \vec{r}, \tag{13.8}
\end{equation*}
where G=6.672(4)xx10^(-11)m^(3)kg^(-1)s^(-2)G=6.672(4) \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2} is the gravitational constant. We also find it useful to deal with a scalar potential energy function U(r)U(r) experienced by mass mm due to mass MM and given by
{:(13.9)U(r)=-(GMm)/(r):}\begin{equation*}
U(r)=-\frac{G M m}{r} \tag{13.9}
\end{equation*}
where vec(F)=- vec(grad)U\vec{F}=-\vec{\nabla} U. The function is U(r)U(r) is proportional to mm, so it is better to think about the potential Phi(r)=U(r)//m\Phi(r)=U(r) / m, the potential energy per unit mass. Newton's law of gravitation can therefore be turned into a field equation straightforwardly. The field version of the law describes the force per unit mass vec(g)( vec(x))\vec{g}(\vec{x}) experienced at a position vec(x)\vec{x} due to the presence of a distribution of mass with density rho( vec(x)^('))\rho\left(\vec{x}^{\prime}\right). The generalization of the force law is
We can then show that (i) the function Phi\Phi obeys the important rule vec(g)( vec(x))=-grad Phi( vec(x))\vec{g}(\vec{x})=-\nabla \Phi(\vec{x}), and, most importantly, (ii) Phi\Phi can be computed from a given distribution of mass. This is the purpose of the next example.
Example 13.1
(i) Using the potential Phi( vec(x))\Phi(\vec{x}) we can write ^(7){ }^{7}
which allows us to take the divergence of vec(g)\vec{g} in eqn 13.10 to find
{:[ vec(grad)*g( vec(x))=-4pi G intd^(3)x^(')rho( vec(x)^('))delta^((3))(( vec(x))- vec(x)^('))],[(13.16)=-4pi G rho( vec(x))]:}\begin{align*}
\vec{\nabla} \cdot g(\vec{x}) & =-4 \pi G \int \mathrm{~d}^{3} x^{\prime} \rho\left(\vec{x}^{\prime}\right) \delta^{(3)}\left(\vec{x}-\vec{x}^{\prime}\right) \\
& =-4 \pi G \rho(\vec{x}) \tag{13.16}
\end{align*}
Since we also have vec(grad)*g( vec(x))=- vec(grad)^(2)Phi\vec{\nabla} \cdot g(\vec{x})=-\vec{\nabla}^{2} \Phi, we conclude that
{:(13.17) vec(grad)^(2)Phi( vec(x))=4pi G rho( vec(x)):}\begin{equation*}
\vec{\nabla}^{2} \Phi(\vec{x})=4 \pi G \rho(\vec{x}) \tag{13.17}
\end{equation*}
This is the gravitational version of Poisson's equation. ^(9){ }^{9}
where vec(g)\vec{g} is the gravitational field. The law can be expressed in terms of the potential Phi\Phi (via vec(g)=- vec(grad)Phi\vec{g}=-\vec{\nabla} \Phi ). The potential field is the solution of Poisson's equation ^(9){ }^{9} Recall that Laplace's equation for a field in the absence of sources says vec(grad)^(2)Phi=0\vec{\nabla}^{2} \Phi=0, but Poisson's equation for the field in the presence of sources says vec(grad)^(2)Phi=4pi G rho\vec{\nabla}^{2} \Phi=4 \pi G \rho, where 4pi G rho4 \pi G \rho is the density of the source. ^(10){ }^{10} Note that eqn 13.18 is the same as presented in eqn 0.12 , and eqn 13.19 is the same as presented in eqn 0.15 . ^(7){ }^{7} Proof: Check by a direct computation that
{:(13.19) vec(grad)^(2)Phi( vec(x))=4pi G rho( vec(x)).:}\begin{equation*}
\vec{\nabla}^{2} \Phi(\vec{x})=4 \pi G \rho(\vec{x}) . \tag{13.19}
\end{equation*}
^(11){ }^{11} The idea of the Green's function is that for a differential equation hat(L)A(x)=\hat{L} A(x)=f(x)f(x), with hat(L)\hat{L} representing a differential operator, we define the Green's function as hat(L)G(x,y)=delta(x-y)\hat{L} G(x, y)=\delta(x-y). This is useful, as the Green's function can then be used to build a solution using the prescription
{:(13.20)A(x)=intdyG(x","y)f(y):}\begin{equation*}
A(x)=\int \mathrm{d} y G(x, y) f(y) \tag{13.20}
\end{equation*}
^(12){ }^{12} We know that both the energy density and metric field are tensor fields, so we might expect Newton's law to be equivalent to one component of the Einstein equation.
Fig. 13.2 The relationship between the source of energy-momentum TT and the metric field gg in the Einstein equation. The source is constrained by conservation laws (via grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0 ) affecting its coupling to the metric field that, in turn, determines the geometry. The metric field also determines the form of TT itself. ^(13){ }^{13} We can be a little more specific here. In electromagnetism (another linear theory), we often specify the currents and charges and then solve Maxwell's equations to find the electromagnetic fields. The same can be done in Newtonian gravity: specify the mass distribution and solve for Phi\Phi. The same cannot be done in general relativity since the source of gravitation is the energy-momentum tensor T\boldsymbol{T}, which is described in terms of the metric g\boldsymbol{g}, precisely the field we are trying to find. This forces us to attempt to find consistent values of g\boldsymbol{g} and T\boldsymbol{T} simultaneously, which is very difficult.
Note the form of this equation that says that two derivatives of the potential field are proportional to the mass density rho\rho, which is the source of the field. We therefore have a field theory of Newtonian gravitation that deals in a scalar field Phi( vec(x),t)\Phi(\vec{x}, t). We input a position in spacetime and output a function Phi\Phi. This can be used to compute the gravitational field and hence the force on any mass.
Example 13.2
Green's functions are a very useful tool in solving equations like the Poisson equation. ^(11){ }^{11} For our problem, we seek a solution to the equation
We can then use the resulting Green's function G(( vec(x)), vec(x)^('))G\left(\vec{x}, \vec{x}^{\prime}\right) to build a solution to Poisson's equation using the prescription
We shall try to fashion our theory of gravitation following the pattern set by Newton's field theory. In place of the Newtonian potential, we shall substitute the metric. In place of the mass density, we substitute the energy density. If we postulate that the Einstein equation has the same form as Poisson's equation for the Newtonian potential field, then Einstein's equation must look like
{:(13.24)(del^(2)g)=kappa T:}\begin{equation*}
\left(\partial^{2} g\right)=\kappa T \tag{13.24}
\end{equation*}
where kappa\kappa is a constant. ^(12){ }^{12} That is to say that two derivatives of the metric field gg are proportional to the energy density TT.
Although these steps might make us optimistic, there must also be differences between Einstein and Newton's pictures of gravitation. First, Newton's law is instantaneous, running counter to the rules of special relativity. In addition, Fermat's principle says that light propagates in straight lines in flat spacetime but experiments show the bending of light from stars by gravitation, implying that gravity must warp spacetime. This means that, for Einstein's gravity, the geometry of curved space is caused by the energy density of matter which acts as a source (just like rho\rho is a source of Phi)\Phi). However, the arrangement of energy density is itself determined by the geometry encoded in the metric field, leading to the situation shown in Fig. 13.2. We conclude that, in contrast to Newton, Einstein's gravitation is necessarily nonlinear: gravitational energy is acted on by gravity itself. ^(13){ }^{13}
We have now reached the point where we must fit the geometrical and physical fields together. The key physical principles determining the content on the right-hand side of a relativistic equation of motion are as follows. (i) Local causality: A signal can be sent between two points if, and only if, their separation is not spacelike. Colloquially: nothing travels faster than light. Our fields must therefore be consistent with special relativity. (ii) Local conservation of energy and momentum: All matter fields in the Universe have energy-momentum. This energy-momentum is locally conserved. That is to say, it must obey a continuity equation.
We also require energy to obey another condition. (iii) The dominant energy condition says that to any observer the local energy density appears non-negative and the local energy flow vector is not spacelike (i.e. it is timelike or null). We therefore demand that positive energy, which cannot move around a system at a speed faster than that of light, is locally conserved. ^(14){ }^{14}
All of the physical principles relating to the behaviour of mass energy can be embodied in an energy-momentum tensor field T(x)\boldsymbol{T}(x), as described in the previous chapter. This tensor must describe positive energy and encodes local conservation of energy-momentum (at least approximately) via the constraint grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0. The energy-momentum tensor vanishes only if all of the matter fields vanish. This means that all fields cost energy, and this is the source of gravitation.
In the next section, we shall, finally, attempt to marry the geometry of curvature embodied in R\boldsymbol{R} and the mass energy of the fields of the Universe, expressed in T\boldsymbol{T} in a tensor equation.
13.3 An incorrect guess
The apparently obvious field equation turns out to be incorrect.
The energy-momentum tensor has two slots or, equivalently, its components have two indices. We are therefore tempted to guess that the Einstein equation, that fixes the relationship of curvature (on the left) to the energy density of the matter fields (on the right) has a (2,0)(2,0) object on the left and T^(mu nu)T^{\mu \nu} on the right. The obvious, but incorrect, natural candidate would be that the curvature-defining object is the Ricci tensor, with components R^(mu nu)R^{\mu \nu}. This would give us the field equation
with kappa\kappa a constant. ^(15){ }^{15} But why is this wrong? The key is to remember that we must have the conservation equation grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0. We examine the consequence of this on our candidate field equation by taking the divergence
Since grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0 (or, equivalently T^(mu nu)_(;mu)=0T^{\mu \nu}{ }_{; \mu}=0 ), the right-hand side of eqn 13.26 must vanish. What happens to the geometrical left-hand side? ^(14){ }^{14} There is a less stringent weak energy condition, which says that the energy density measured by any observer must be non-negative.
15 This seems a good idea since the Ricci tensor is a sort of average of R\boldsymbol{R} that retains those parts of curvature that, in a gravitational equation, can cause volumes to shrink. This was Einstein's original suggestion in October 1915. ^(16){ }^{16} We make use of the symmetries of R_(alpha beta gamma delta)R_{\alpha \beta \gamma \delta}, where swaps within the firs and second pairs of indices result in a minus sign. ^(17){ }^{17} Do not confuse the Einstein tensor field G\boldsymbol{G} with the gravitational constant GG. It's unfortunate that they have the same symbol, but context should make clear which is which. In later chapters, we'll use units in which G=1G=1. ^(18){ }^{18} We use the compatibility condition g_(mu nu;alpha)=0g_{\mu \nu ; \alpha}=0 here. ^(19){ }^{19} If you don't see this immediately, the steps are (i) raise the mu\mu index in the second term to find R^(mu nu)_(;mu)-R^(alpha nu)_(;alpha)R^{\mu \nu}{ }_{; \mu}-R^{\alpha \nu}{ }_{; \alpha}; (ii) rename the alpha\alpha index mu\mu and the expression vanishes. ^(20){ }^{20} Remember that the effect of the covariant derivative on a scalar field is simply
grad f=f_(;mu)=f_(,mu)=(del f)/(delx^(mu)).\nabla f=f_{; \mu}=f_{, \mu}=\frac{\partial f}{\partial x^{\mu}} .
To answer this, we need a mathematical result to constrain our equations.
Example 13.3
As we'll discuss in Chapter 43, the components of the Riemann tensor satisfy a geometrical constraint known as the Bianchi identity, whose component form is written as
For now, it is enough to say that this identity comes from a general principle that 'the boundary of a boundary is zero', an idea that leads to the well-known vector identity that the divergence of the curl of a vector field vanishes. We can contract indices in this identity, setting alpha=lambda\alpha=\lambda, to find ^(16){ }^{16}
This equation is known as the contracted Bianchi identity and we will use it below.
To examine the consequence of the identity discussed in the last example, we now define the trace-reversed Ricci tensor, also known as the Einstein tensor field, as the (2,0)(2,0) tensor G(x)\boldsymbol{G}(x) with components ^(17){ }^{17}
Using the contracted Bianchi identity ( R_(;mu)=2R^(alpha)_(mu;alpha)R_{; \mu}=2 R^{\alpha}{ }_{\mu ; \alpha} ) on the second term on the right, we find ^(19){ }^{19}
Although this doesn't look particularly scandalous, it causes a major problem for the divergence of our candidate equation: R^(mu nu)_(;mu)=kappaT^(mu nu)_(;mu)=R^{\mu \nu}{ }_{; \mu}=\kappa T^{\mu \nu}{ }_{; \mu}= 0 . We combine this expression with eqn 13.33 to conclude
The guts of this latter equation ^(20)(R^(;nu)=0){ }^{20}\left(R^{; \nu}=0\right) forces the Ricci scalar field R(x)R(x) to be a constant throughout the Universe. This on its own is not
a problem. However, the trouble comes when we examine the candiate equation (eqn 13.25) again, since we find that
That is, the trace g^(mu nu)T_(mu nu)g^{\mu \nu} T_{\mu \nu} of T_(mu nu)T_{\mu \nu} must be constant throughout the Universe. This is unreasonable! It implies a uniform distribution of matter and energy throughout the Universe which is at odds with our experience, where matter is certainly non-uniform. We are forced to drop the candidate equation.
13.4 Einstein's field equation
The problem encountered in the last section leads to the solution.
On our way to proving that our first attempt at a field equation was inadequate, we identified a new (2,0)(2,0) [or (0,2)](0,2)] tensor G\boldsymbol{G}, which had components G_(mu nu)=R_(mu nu)-(1)/(2)g_(mu nu)RG_{\mu \nu}=R_{\mu \nu}-\frac{1}{2} g_{\mu \nu} R. It is this object that provides the solution to our problem. Therefore, having discarded our first guess, we now consider instead a new candidate field equation ^(21){ }^{21}
We know from the last section that the divergence of this equation is zero on both sides. This, it turns out, is exactly the field equation for which we've been searching. However, we have yet to work out the constant of proportionality kappa\kappa. This can be evaluated using the condition that this equation is compatible with the results of Newtonian gravitation in the limit of weak fields (i.e. with the equation vec(grad)^(2)Phi=4pi G rho\vec{\nabla}^{2} \Phi=4 \pi G \rho ).
Example 13.4
In the weak-field limit, spacetime is only very slightly curved and therefore the metric is very close to the flat Minkowski metric, so g_(00)~~g^(00)~~-1g_{00} \approx g^{00} \approx-1 and g_(ii)~~g^(ii)~~1g_{i i} \approx g^{i i} \approx 1. As a result, we have for any tensor A\boldsymbol{A} in this geometry, that A^(00)~~A_(00)A^{00} \approx A_{00}, along with A^(ii)=A_(ii)A^{i i}=A_{i i}. Moreover, the energy-momentum tensor is dominated by the energy density part (because rho≫p//c^(2)\rho \gg p / c^{2} ) so the only significant element of T\boldsymbol{T} is T^(00)=rhoc^(2)T^{00}=\rho c^{2}, with all other elements vanishingly small. This means that, for all of the spatial (ij) components of the Einstein tensor, we have
from which we find R^(11)=R^(22)=R^(33)=R//2R^{11}=R^{22}=R^{33}=R / 2 and so
{:(13.38)R^(11)+R^(22)+R^(33)=3R//2:}\begin{equation*}
R^{11}+R^{22}+R^{33}=3 R / 2 \tag{13.38}
\end{equation*}
As a result, R=g_(mu nu)R^(mu nu)=-R^(00)+(3)/(2)RR=g_{\mu \nu} R^{\mu \nu}=-R^{00}+\frac{3}{2} R and therefore R=2R^(00)R=2 R^{00}. This allows us to calculate G^(00)=R^(00)-(1)/(2)g^(00)RG^{00}=R^{00}-\frac{1}{2} g^{00} R which gives
Recall that, with the weak-field metric, the only appreciable connection coefficients are (replacing factors of c) Gamma_(00)^(i)=del(Phi//c^(2))//delx^(i)\Gamma_{00}^{i}=\partial\left(\Phi / c^{2}\right) / \partial x^{i} and so R^(i)_(0j0)=del^(2)(Phi//c^(2))//delx^(i)delx^(j)R^{i}{ }_{0 j 0}=\partial^{2}\left(\Phi / c^{2}\right) / \partial x^{i} \partial x^{j}. We can then read off ^(22){ }^{22} that R_(00)=(1//c^(2)) vec(grad)^(2)PhiR_{00}=\left(1 / c^{2}\right) \vec{\nabla}^{2} \Phi and so
^(21){ }^{21} In case this mathematics is a distraction, here's a physics recap. The Ricci tensor represents an average of the Riemann curvature tensor that will encode those parts of curvature that cause volumes to shrink owing to the gravitational interaction. However, equating the Ricci tensor to the energying the Ricci tensor to the energy-
momentum tensor gives a theory that momentum tensor gives a theory that
doesn't conserve energy (i.e. it is incomdoesn't conserve energy (i.e. it is incom-
patible with grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0 ). The Bianchi identity (geometrically) encodes the desired energy conservation. Using a trace-reversed Ricci tensor, known as the Einstein tensor, builds energy conservation back into the theory since, via the Bianchi identity, it allows the resulting equations to conserve energy.
In this example, we will temporarily reinsert the factors of cc so it's a bit more clear how large various factors are.
{:[^(22)" Explicitly we have "R_(00)=],[sum_(i=1)^(3)R_(0i0)^(i)". "]:}\begin{aligned}
& { }^{22} \text { Explicitly we have } R_{00}= \\
& \sum_{i=1}^{3} R_{0 i 0}^{i} \text {. }
\end{aligned}
Putting this together with T^(00)=rhoc^(2)T^{00}=\rho c^{2}, the Einstein equation G=kappa T\boldsymbol{G}=\kappa \boldsymbol{T} predicts a universal field equation of
This can be compared with Poisson's equation for gravitation (the field equation that corresponds to Newton's universal law of gravitation), which reads vec(grad)^(2)Phi=4pi G rho\vec{\nabla}^{2} \Phi=4 \pi G \rho. Hence, we deduce that
In SI units, the factor is kappa=8pi G//c^(4)~~2.08 xx10^(-43)N^(-1)\kappa=8 \pi G / c^{4} \approx 2.08 \times 10^{-43} \mathrm{~N}^{-1}.
The result of the previous example, setting c=1c=1 again, is that kappa=8pi G\kappa=8 \pi G. We are now in a position to write a field equation for gravity, known as ^(23){ }^{23} This is the equation Einstein presented in November 1915, after having rejected his first attempt.
the Einstein field equation. ^(23){ }^{23}
The Einstein field equation, version 1
In coordinates,
Let's now consider the trace TT of the energy-momentum tensor T=g^(mu nu)T_(mu nu)T=g^{\mu \nu} T_{\mu \nu}, and hence eqn 13.45 gives (using g^(mu nu)g_(mu nu)=4g^{\mu \nu} g_{\mu \nu}=4 and R=g_(mu nu)R^(mu nu)R=g_{\mu \nu} R^{\mu \nu} ) R-2R=-R=8pi GTR-2 R=-R=8 \pi G T This implies we can also write
{:(13.46)R_(mu nu)+(1)/(2)g_(mu nu)(8pi GT)=8pi GT_(mu nu):}\begin{equation*}
R_{\mu \nu}+\frac{1}{2} g_{\mu \nu}(8 \pi G T)=8 \pi G T_{\mu \nu} \tag{13.46}
\end{equation*}
The previous example therefore leads to a second form for the Einstein field equation:
Our method of arriving at the Einstein equation has been rather haphazard. We might ask if we can add other terms to the right-hand side of the Einstein equation. In fact, it is permissible to add any scalar multiple of the metric field tensor, so we can take our (version 1) equation G(x)=8pi GT(x)\boldsymbol{G}(x)=8 \pi G \boldsymbol{T}(x) and put an extra -Lambda g(x)-\Lambda \boldsymbol{g}(x) on the right-hand side and if Lambda\Lambda is just a constant then our new equation will still work. In fact, the property that grad_(alpha)g=g_(mu nu;alpha)=0\nabla_{\alpha} \boldsymbol{g}=g_{\mu \nu ; \alpha}=0 means that the divergence arguments
we have been using won't be troubled by this extra term. We therefore write our third version of the Einstein equation as follows:
The constant of proportionality Lambda\Lambda is known as the cosmological constant. Physically, the cosmological constant represents an extra source of energy in the Universe, in addition to the matter fields. Owing to the difference in sign compared to positive T_(mu nu)T_{\mu \nu} (as demanded by the dominant energy condition), a positive value of Lambda\Lambda will give a repulsive gravitational interaction.
In order to retain the simplicity of the equation G=8pi GT\boldsymbol{G}=8 \pi G \boldsymbol{T}, we could treat the cosmological constant term as if it were a source of energymomentum and write
is a contribution to the energy-momentum tensor from sources other than the matter fields. The introduction and rejection of Lambda\Lambda by Einstein has become a famous story in physics folklore. We shall discuss its consequences in Part III of the book. ^(24){ }^{24}
We have arrived at an equation of motion for the metric field that determines gravitation. Our task is now to explore its consequences on our Universe. In the coming chapters, it is useful to keep in mind a few helpful heuristics we have seen already in dealing with general relativity.
A metric is visualized via its light cones. These are found using ds^(2)=0\mathrm{d} s^{2}=0 along null geodesics.
We use two types of reference frame: (i) a coordinate frame, where our geometric arguments are often simplest; (ii) an orthonormal frame, where observers make measurements.
Coordinates have no intrinsic metric significance but, in cases of spherical symmetry, useful coordinates to have in mind are spherical polars (t,r,theta,phi)(t, r, \theta, \phi).
Indices are raised/lowered in the coordinate frame using the metric. Indices in orthonormal frames are manipulated with the Minkowski tensor. The connection coefficients don't vanish in the orthonormal frame. If we want this we need to shift to a local inertial frame. ^(24){ }^{24} A brief history of solutions involving Lambda\Lambda is given in Chapter 18.
General covariance says that a valid tensor equation in flat space is a valid tensor equation in curved space. To upgrade tensor equations, exchange eta rarr g\boldsymbol{\eta} \rightarrow \boldsymbol{g}. To upgrade derivatives, use the comma goes to semicolon rule.
When manipulating vectors, it's often helpful to use (i) g_(mu nu)=g_{\mu \nu}=e_(mu)*e_(nu)\boldsymbol{e}_{\mu} \cdot \boldsymbol{e}_{\nu}; (ii) g_(mu nu;alpha)=0g_{\mu \nu ; \alpha}=0. For timelike velocities u*u=-1\boldsymbol{u} \cdot \boldsymbol{u}=-1.
The perfect fluid, in the orthonormal frame for spherical polars, has components
The left-hand side of the Einstein field equation encodes geometry using the Ricci tensor and scalar to form the Einstein tensor G\boldsymbol{G} with components
The right-hand side of Einstein's equation encodes the energymomentum of matter fields using the tensor T\boldsymbol{T}.
These are tied together in Einstein's equation, subject to the key constraint that energy-momentum is locally conserved. The Einstein equation is G(x)=8pi GT(x)-Lambda g(x)\boldsymbol{G}(x)=8 \pi G \boldsymbol{T}(x)-\Lambda \boldsymbol{g}(x) and has components
where r=| vec(x)|r=|\vec{x}|. We shall prove this. (a) Evaluate (del)/(delx^(2))r\frac{\partial}{\partial x^{2}} r and (del)/(delx^(2))r^(-1)\frac{\partial}{\partial x^{2}} r^{-1}. (b) Show that
where n^(i)=x^(i)//rn^{i}=x^{i} / r. (c) Use the previous result to show that vec(grad)^(2)(r^(-1))=0\vec{\nabla}^{2}\left(r^{-1}\right)=0 for | vec(x)|!=0|\vec{x}| \neq 0. (d) Defining vec(j)= vec(grad)(r^(-1))\vec{j}=\vec{\nabla}\left(r^{-1}\right), use Gauss' theorem to show that
where VV is a volume that encloses vec(x)=0\vec{x}=0. (e) Use the previous result to complete the proof.
(13.3) Verify that taking the trace Tr(G^(mu nu))\operatorname{Tr}\left(G^{\mu \nu}\right) of the Einstein tensor with components G^(mu nu)G^{\mu \nu} gives Tr(G^(mu nu))=-R\operatorname{Tr}\left(G^{\mu \nu}\right)=-R, where RR is the trace of the Ricci tensor.
The momentum flux in, say, the 
